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Question 13 of 25 Two samples of water are mixed together. The first sample has a mass of 1.0 kg and is at 0°C. The second sample has a mass of 0.50 kg and is at 100°C. What is the equilibrium temperature of the water, assuming this is a closed system? (Gwater = 4.18 kJ/kg:°C)​

Sagot :

Answer:

Q = k M T    where k is the absorption constant, Q the heat absorbed, M the mass involved and T the temperature change (for change in temperature)

Q1 = 4.18 kJ/kg * 1 kg * 273 deg K     heat contained

Q2 = 4.18 kJ/kg * .5 kg * 373 deg K     heat contained

Q1 = 1141 kJ deg K

Q2 = 780 kJ deg K

Q = Q1 + Q2 = 1921 kJ deg K

T = 1921 / (4.18 * 1.5) deg K = 306 deg K = (306 - 273) = 33 deg Celsius

(The total mass  is now 1.5 kg)

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